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3 years ago

Two sides of an isosceles triangle are 4 and 8. Which of the following choices could be the measure of the third side?

Mathematics

1 answer:

Feliz [49]3 years ago

4 0

Answer:

8 is answer. isoceles triangles must have two equal sides, where the smaller makes the base and the two larger are the legs.

Step-by-step explanation:

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Solve the system of inequalities to decide if the point (-3,2) is part of the solution; y<=-4x-3, +8y>=7

inna [77]

Answer:

$(x,y) = (-3,2)$ is a solution

Step-by-step explanation:

Given

$y \le -4x + 3$

$x + 8y \ge 7$

Required

Determine if $(x,y) = (-3,2)$ is a solution

$y \le -4x + 3$ becomes

$2 \le -4 * -3 + 3$

$2 \le 15$ --- this is true

$x + 8y \ge 7$

$-3+ 8 * 2 \ge 7$

$13 \ge 7$ --- this is also true

Hence,

$(x,y) = (-3,2)$ is a solution

6 0

3 years ago

Elsies math class has 11 female students and 10 male students. Write the ratio of female students to total number of students in

Art [367]

Answer:

Step-by-step explanation:

Download pdf

3 0

2 years ago

Help please! Step by step!

Ad libitum [116K]

Answer:

0.3

Step-by-step explanation:

Find the number in the tenth place

and look one place to the right for the rounding digit

. Round up if this number is greater than or equal to

and round down if it is less than

0.3

8 0

3 years ago

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.

The inner surface: $Si=4\pi (0.06m)^{2} = 0.04 m^{2}$

The outer surface: $S=4\pi (0.08m)^{2}=0.08m^{2}$

Now we can calculate the charges,

Inner charge: $Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC$

Outer charge: $Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC$

Then, we are able to calculate both fields:

Inner field: $Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}$

Outer field: $Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}$

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0

3 years ago

If A=B prove AUB=AnB

expeople1 [14]

If two sets are same let a and bthen we can say a= b
according to definition of intersection it is function that contains common enements from both setsas a=b so all elemnts are commonso a= a n bor b = a n b
in union the function has all elements of both sets and in set no element repeats so
a u b = aor a u b = b
as a u b = a or b and a n b = a or b so we can conclude here that
a u b = a n b as a = b // defined above

6 0

3 years ago